Question: Simplify and expand the following expression: $ \dfrac{2x + 9}{x + 10}-\dfrac{x}{x - 1} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(x + 10)(x - 1)$ Multiply the first term by $\dfrac{x - 1}{x - 1}$ $ \begin{align*} \dfrac{2x + 9}{x + 10} \times \dfrac{x - 1}{x - 1} & = \dfrac{(2x + 9)(x - 1)}{(x + 10)(x - 1)} \\ & = \dfrac{2x^2 + 7x - 9}{(x + 10)(x - 1)}\end{align*} $ Multiply the second term by $\dfrac{x + 10}{x + 10}$ $ \begin{align*} \dfrac{x}{x - 1} \times \dfrac{x + 10}{x + 10} & = \dfrac{(x)(x + 10)}{(x - 1)(x + 10)} \\ & = \dfrac{x^2 + 10x}{(x - 1)(x + 10)}\end{align*} $ Now we have: $ = \dfrac{2x^2 + 7x - 9}{(x + 10)(x - 1)} - \dfrac{x^2 + 10x}{(x - 1)(x + 10)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2x^2 + 7x - 9 - (x^2 + 10x)}{(x + 10)(x - 1)} $ $ = \dfrac{2x^2 + 7x - 9 - x^2 - 10x}{(x + 10)(x - 1)} $ $ = \dfrac{x^2 - 3x - 9}{(x + 10)(x - 1)}$ Expand the denominator: $ = \dfrac{x^2 - 3x - 9}{x^2 + 9x - 10}$